ib questionbank test€¦ · web view[1 mark] marks. cheme (a1) (c1) notes: accept alternative...

6.1

1a. [2 marks]

Markscheme

OR OR (A1)(A1) (C2)

Note: Award (A1) for −2 and (A1) for completely correct mathematical notation, including weak

inequalities. Accept .

[2 marks]

Examiners report

[N/A]

1b. [2 marks]

Markscheme

–1 and 1.52 (1.51839…) (A1)(A1) (C2)

Note: Award (A1) for −1 and (A1) for 1.52 (1.51839).

[2 marks]

Examiners report

[N/A]

1c. [2 marks]

Markscheme

OR . (A1)(ft)(A1)(ft) (C2)

Note: Award (A1)(ft) for both critical values in inequality or range statements such

as .

Award the second (A1)(ft) for correct strict inequality statements used with their critical values. If an

incorrect use of strict and weak inequalities has already been penalized in (a), condone weak

inequalities for this second mark and award (A1)(ft).

1

[2 marks]

Examiners report

[N/A]

2a. [2 marks]

Markscheme

(M1)

Note: Award (M1) for correct substitution of x = 4 and y = 2 into the function.

k = 3 (A1) (G2)

[2 marks]

Examiners report

[N/A]

2b. [3 marks]

Markscheme

(A1)(A1)(A1)(ft) (G3)

Note: Award (A1) for −48 , (A1) for x−2, (A1)(ft) for their 6x. Follow through from part (a). Award at

most (A1)(A1)(A0) if additional terms are seen.

[3 marks]

Examiners report

[N/A]

2c. [3 marks]

Markscheme

(M1)

Note: Award (M1) for equating their part (b) to zero.

2

x = 2 (A1)(ft)

Note: Follow through from part (b). Award (M1)(A1) for seen.

Award (M0)(A0) for x = 2 seen either from a graphical method or without working.

(M1)

Note: Award (M1) for substituting their 2 into their function, but only if the final answer is −22.

Substitution of the known result invalidates the process; award (M0)(A0)(M0).

−22 (AG)

[3 marks]

Examiners report

[N/A]

2d. [2 marks]

Markscheme

0.861 (0.860548…), 3.90 (3.90307…) (A1)(ft)(A1)(ft) (G2)

Note: Follow through from part (a) but only if the answer is positive. Award at most (A1)(ft)(A0) if

answers are given as coordinate pairs or if extra values are seen. The function f (x) only has two x-

intercepts within the domain. Do not accept a negative x-intercept.

[2 marks]

Examiners report

[N/A]

2e. [4 marks]

Markscheme

3

(A1)(A1)(ft)(A1)(ft)(A1)(ft)

Note: Award (A1) for correct window. Axes must be labelled.

(A1)(ft) for a smooth curve with correct shape and zeros in approximately correct positions relative to

each other.

(A1)(ft) for point P indicated in approximately the correct position. Follow through from their x-

coordinate in part (c). (A1)(ft) for two x-intercepts identified on the graph and curve reflecting

asymptotic properties.

[4 marks]

Examiners report

[N/A]

3a. [4 marks]

Markscheme

4

(A1)(A1)(A1)(A1)

Note: Award (A1) for correct window (condone a window which is slightly off) and axes labels. An

indication of window is necessary. −1 to 3 on the x-axis and −2 to 12 on the y-axis and a graph in that

window.

(A1) for correct shape (curve having cubic shape and must be smooth).

(A1) for both stationary points in the 1st quadrant with approximate correct position,

(A1) for intercepts (negative x-intercept and positive y intercept) with approximate correct position.

[4 marks]

Examiners report

[N/A]

3b. [1 mark]

Markscheme

Rick (A1)

Note: Award (A0) if extra names stated.

[1 mark]

Examiners report

5

[N/A]

3c. [2 marks]

Markscheme

2(1)3 − 9(1)2 + 12(1) + 2 (M1)

Note: Award (M1) for correct substitution into equation.

= 7 (A1)(G2)

[2 marks]

Examiners report

[N/A]

3d. [3 marks]

Markscheme

6x2 − 18x + 12 (A1)(A1)(A1)

Note: Award (A1) for each correct term. Award at most (A1)(A1)(A0) if extra terms seen.

[3 marks]

Examiners report

[N/A]

3e. [2 marks]

Markscheme

6x2 − 18x + 12 = 0 (M1)

6

Note: Award (M1) for equating their derivative to 0. If the derivative is not explicitly equated to 0, but a

subsequent solving of their correct equation is seen, award (M1).

6(x − 1)(x − 2) = 0 (or equivalent) (M1)

Note: Award (M1) for correct factorization. The final (M1) is awarded only if answers are clearly

stated.

Award (M0)(M0) for substitution of 1 and of 2 in their derivative.

x = 1, x = 2 (AG)

[2 marks]

Examiners report

[N/A]

3f. [3 marks]

Markscheme

6 < k < 7 (A1)(A1)(ft)(A1)

Note: Award (A1) for an inequality with 6, award (A1)(ft) for an inequality with 7 from their part (c)

provided it is greater than 6, (A1) for their correct strict inequalities. Accept ]6, 7[ or (6, 7).

[3 marks]

Examiners report

[N/A]

4a. [2 marks]

Markscheme

(A2) (C2)

Note: Accept equivalent notation. Award (A1)(A0) for .

Award (A1) for a clear statement that demonstrates understanding of the meaning of domain. For

example, should be awarded (A1)(A0).

7

[2 marks]

Examiners report

[N/A]

4b. [1 mark]

Markscheme

(A1) (C1)

Note: The command term “Draw” states: “A ruler (straight edge) should be used for straight lines”; do

not accept a freehand line.

[1 mark]

Examiners report

[N/A]

4c. [1 mark]

Markscheme

2 (A1)(ft) (C1)

Note: Follow through from part (b)(i).

8

[1 mark]

Examiners report

[N/A]

4d. [2 marks]

Markscheme

(A1)(A1) (C2)

Note: Award (A1) for both end points correct and (A1) for correct strict inequalities.

Award at most (A1)(A0) if the stated variable is different from or for example is (A1)

(A0).

[2 marks]

Examiners report

[N/A]

5a. [1 mark]

Markscheme

400 (USD) (A1) (C1)

[1 mark]

Examiners report

[N/A]

5b. [2 marks]

Markscheme

(M1)

9

Note: Award (M1) for equating to or for comparing the difference

between the two expressions to zero or for showing a sketch of both functions.

(A1) (C2)

Note: Accept 9 months.

[2 marks]

Examiners report

[N/A]

5c. [3 marks]

Markscheme

(M1)(M1)

Note: Award (M1) for correct substitution of into equation for , (M1) for finding the

difference between a value/expression for and a value/expression for . The first (M1) is implied if

7671.25 seen.

4870 (USD) (4871.25) (A1) (C3)

Note: Accept 4871.3.

[3 marks]

10

Examiners report

[N/A]

6a. [2 marks]

Markscheme

(A1)(A1) (C2)

Note: Award (A1) for constant, (A1) for the constant being 3.

The answer must be an equation.

[2 marks]

Examiners report

[N/A]

6b. [2 marks]

Markscheme

(M1)

Note: Award (M1) for correct substitution into axis of symmetry formula.

OR

(M1)

Note: Award (M1) for correctly differentiating and equating to zero.

OR11

(or equivalent)

(or equivalent) (M1)

Note: Award (M1) for correct substitution of and in the original quadratic function.

(A1)(ft) (C2)

Note: Follow through from part (a).

[2 marks]

Examiners report

[N/A]

6c. [2 marks]

Markscheme

OR (A1)(A1)

Note: Award (A1) for two correct interval endpoints, (A1) for left endpoint excluded and right

endpoint included.

[2 marks]

Examiners report

[N/A]

7a. [2 marks]

Markscheme12

(or equivalent) (R1)

Note: For (R1) accept substitution of or into the equation followed by a confirmation

that or .

(since the point satisfies the equation of the line,) A lies on (A1)

Note: Do not award (A1)(R0).

[2 marks]

Examiners report

[N/A]

7b. [2 marks]

Markscheme

OR seen (M1)

Note: Award (M1) for at least one correct substitution into the midpoint formula.

(A1)(G2)

Notes: Accept .

Award (M1)(A0) for .

Award (G1) for each correct coordinate seen without working.

13

[2 marks]

Examiners report

[N/A]

7c. [2 marks]

Markscheme

(M1)

Note: Award (M1) for a correct substitution into distance between two points formula.

(A1)(G2)

[2 marks]

Examiners report

[N/A]

7d. [5 marks]

Markscheme

gradient of (M1)

Note: Award (M1) for correct substitution into gradient formula.

(A1)

Note: Award (M1)(A1) for gradient of with or without working

14

gradient of the normal (M1)

Note: Award (M1) for the negative reciprocal of their gradient of AC.

OR (M1)

Note: Award (M1) for substitution of their point and gradient into straight line formula. This (M1)

can only be awarded where (gradient) is correctly determined as the gradient of the normal to AC.

OR OR (A1)

Note: Award (A1) for correctly removing fractions, but only if their equation is equivalent to the

given equation.

(AG)

Note: The conclusion must be seen for the (A1) to be awarded.

Where the candidate has shown the gradient of the normal to , award (M1) for

and (A1) for (therefore) .

Simply substituting into the equation of with no other prior working, earns no marks.

[5 marks]

Examiners report15

[N/A]

7e. [2 marks]

Markscheme

(A1)(A1)(G2)

Note: Award (A1) for 6, (A1) for 6.5. Award a maximum of (A1)(A0) if answers are not given as a

coordinate pair. Accept .

Award (M1)(A0) for an attempt to solve the two simultaneous equations and

algebraically, leading to at least one incorrect or missing coordinate.

[2 marks]

Examiners report

[N/A]

7f. [1 mark]

Markscheme

3.3541 (A1)

Note: Answer must be to 5 significant figures.

[1 mark]

Examiners report

[N/A]

7g. [3 marks]

Markscheme

16

(M1)(M1)

Notes: Award (M1) for correct substitution into area of triangle formula.

If their triangle is a quarter of the rhombus then award (M1) for multiplying their triangle by 4.

If their triangle is a half of the rhombus then award (M1) for multiplying their triangle by 2.

OR

(M1)(M1)

Notes: Award (M1) for doubling MD to get the diagonal BD, (M1) for correct substitution into the

area of a rhombus formula.

Award (M1)(M1) for their (f).

(A1)(ft)(G3)

Notes: Follow through from parts (c) and (f).

[3 marks]

Examiners report

[N/A]

8a. [2 marks]

Markscheme

17

(M1)

Note: Award (M1) for correct substitution into equation.

(A1) (C2)

Examiners report

Question 14: Exponential Function

This question was well-answered by the majority of candidates.

Part (a) was generally accessible, unless subtraction was used in the rearrangement of the formula.

8b. [2 marks]

Markscheme

(M1)

Note: Award (M1) for correct substitution into formula. Follow through from part (a).

(A1)(ft) (C2)

Note: The answer must be an integer.

Examiners report

Part (b) required an integer value.

8c. [2 marks]

Markscheme

(M1)

(A1)(ft) (C2)

Note: Award (M1) for setting up the equation. Accept alternative methods such as , or a

sketch of and with indication of point of intersection. Follow through

from (a).

Examiners report

18

Part (c) saw good use of the GDC – with many sketch graphs being shown on paper (this is to be

encouraged) – as well as attempts using logarithms. Use of the GDC is to be encouraged as its efficient

use is a mandatory part of the course. Logarithms are not discouraged – but they are not a necessary

component of the course and it is easy to construct equations that are not accessible to solution by

logarithms.

9a. [3 marks]

Markscheme

i) (A1)

ii) (A1)(A1) (C3)

Note: Award (A1) for “ ” and (A1) for “ ” seen as part of an equation.

Examiners report

Question 8: Rational function.

Few candidates could find the -intercept of the rational function. Many candidates did appreciate that

the curve does not cross the asymptote. Often the candidates wrote down the equation of the horizontal

asymptote rather than the equation of the vertical asymptote. The most frequent incorrect sketch was

that of suggesting that the candidate did not understand that the curve is not

linear and had taken insufficient care in entering the function into the calculator. Some candidates that

appreciated the shape of the curve did not earn marks on account of the poor quality of their sketches,

which either crossed, or veered away from, the asymptotes.

9b. [3 marks]

Markscheme

19

(A1)(ft)(A1)(ft)(A1) (C3)

Note: Award (A1)(ft) for correct -intercept, (A1)(ft) for asymptotic behaviour at -axis, (A1) for

approximately correct shape (cannot intersect the horizontal asymptote of ). Follow through

from part (a).

Examiners report

Question 8: Rational function.

Few candidates could find the -intercept of the rational function. Many candidates did appreciate that

the curve does not cross the asymptote. Often the candidates wrote down the equation of the horizontal

asymptote rather than the equation of the vertical asymptote. The most frequent incorrect sketch was

that of suggesting that the candidate did not understand that the curve is not

linear and had taken insufficient care in entering the function into the calculator. Some candidates that

20

appreciated the shape of the curve did not earn marks on account of the poor quality of their sketches,

which either crossed, or veered away from, the asymptotes.

10a. [1 mark]

Markscheme

(A1)

Examiners report

Question 3: Quadratic function, problem solving.

Parts (a) (finding the initial height) and (b) (finding the height after 15 seconds), were done very well

by the majority of candidates. Many struggled to translate question (c) to find the (positive) zeros of the

function, or did not write that down, losing a possible method mark. The derivative in part (d) was no

problem for most; only very few used instead of . The maximum height reached was calculated

correctly by the majority of candidates, but many lost the mark in part (e)(i) as they simply substituted

40 into their derivative or calculated the height at points close to 40. Only a few candidates showed

correct method for part (f). Several were still able to obtain 2 marks as a result of “trial and error” of

integer values for . Some candidate seem to have a problem with the notation “ ”, where this

is interpreted as , resulting in incorrect answers throughout.

10b. [2 marks]

Markscheme

(M1)

Note: Award (M1) for substitution of in expression for .

(A1)(G2)

Examiners report







21





10c. [2 marks]

Markscheme

(M1)

Note: Award (M1) for setting to zero.

(A1)(G2)

Note: Award at most (M1)(A0) for an answer including .

Award (A0) for an answer of without working.

Examiners report











10d. [2 marks]

Markscheme

(A1)(A1)

Note: Award (A1) for , (A1) for . Award at most (A1)(A0) if extra terms seen. Do not accept

for .

22

Examiners report











10e. [3 marks]

Markscheme

i) (M1)

Note: Award (M1) for setting their derivative, from part (d), to zero, provided the correct conclusion is

stated and consistent with their .

OR

(M1)

Note: Award (M1) for correct substitution into axis of symmetry formula, provided the correct

conclusion is stated.

(AG)

ii) (M1)

Note: Award (M1) for substitution of in expression for .

(A1)(G2)

23

Examiners report











10f. [3 marks]

Markscheme

(M1)

Note: Award (M1) for setting to . Accept inequality sign.

OR

24

M1

Note: Award (M1) for correct sketch. Indication of scale is not required.

(A1)

Note: Award (A1) for and seen.

(total time ) (A1)(G2)

Note: Award (G1) if the two endpoints are given as the final answer with no working.

Examiners report








25




11a. [3 marks]

Markscheme

(i) 2p + q = 11 and 4p + q = 17 (M1)

Note: Award (M1) for either two correct equations or a correct equation in one unknown equivalent to

2p = 6 .

p = 3 (A1)

(ii) q = 5 (A1) (C3)

Notes: If only one value of p and q is correct and no working shown, award (M0)(A1)(A0).

[3 marks]

Examiners report

Candidates both understood how to interpret a mapping diagram correctly and did very well on this

question or the question was very poorly answered or not answered at all. Writing down two

simultaneous equations in part (a) proved to be elusive to many and this prevented further work on

this question. Candidates who were able to find values of p and q (correct or otherwise) invariably

made a good attempt at finding the value of s in part (c).

11b. [1 mark]

Markscheme

r = 8 (A1)(ft) (C1)

Note: Follow through from their answers for p and q irrespective of whether working is seen.

26

[1 mark]

Examiners report






11c. [2 marks]

Markscheme

3 × 2s + 5 = 197 (M1)

Note: Award (M1) for setting the correct equation.

s = 6 (A1)(ft) (C2)

Note: Follow through from their values of p and q.

[2 marks]

Examiners report






12a. [3 marks]

Markscheme

(i) (M1)

Note: Award (M1) for correct substitution in formula.

27

OR

(M1)

Note: Award (M1) for setting up 2 correct simultaneous equations.

OR

(M1)

Note: Award (M1) for correct derivative of equated to zero.

(A1) (C2)

(ii)

(A1)(ft) (C1)

Note: Follow through from their value for b.

Note: Alternatively candidates may answer part (a) using the method below, and not as two separate

parts.

(M1)

(A1)

(A1) (C3)

28

[3 marks]

Examiners report

Question 11 proved to be the most problematic of the whole paper. Many candidates attempted this

question but were not able to set up a system of equations to find the value of b or use the formula

. From the working seen, many candidates did not understand the non-standard notation for

the domain, with a number believing it to be a coordinate pair. This was taken into careful

consideration by the senior examiners when setting the grade boundaries for this paper.

12b. [3 marks]

Markscheme

–5 ≤ y ≤ 4 (A1)(ft)(A1)(ft)(A1) (C3)

Notes: Accept [–5, 4]. Award (A1)(ft) for –5, (A1)(ft) for 4. (A1) for inequalities in the correct direction

or brackets with values in the correct order or a clear word statement of the range. Follow through

from their part (a).

[3 marks]

Examiners report

Question 11 proved to be the most problematic of the whole paper. Many candidates attempted this

question but were not able to set up a system of equations to find the value of b or use the formula

. From the working seen, many candidates did not understand the non-standard notation for

the domain, with a number believing it to be a coordinate pair. This was taken into careful

consideration by the senior examiners when setting the grade boundaries for this paper.

13a. [4 marks]

Markscheme

29

(A1) for indication of window and labels. (A1) for smooth curve that does not enter the first quadrant,

the curve must consist of one line only.

(A1) for x and y intercepts in approximately correct positions (allow ±0.5).

(A1) for local maximum and minimum in approximately correct position. (minimum should be 0 ≤ x ≤ 1

and –2 ≤ y ≤ –4 ), the y-coordinate of the maximum should be 0 ± 0.5. (A4)

[4 marks]

Examiners report30

This question caused the most difficulty to candidates for two reasons; its content and perhaps lack of

time.

Drawing/sketching graphs is perhaps the area of the course that results in the poorest responses. It is

also the area of the course that results in the best. It is therefore the area of the course that good

teaching can influence the most.

Candidates should:

● Use the correct scale and window. Label the axes.

● Enter the formula into the GDC and use the table function to determine the points to be plotted.

● Refer to the graph on the GDC when drawing the curve.

● Draw a curve rather than line segments; ensure that the curve is smooth.

● Use a pencil rather than a pen so that required changes once further information has been

gathered (the turning points, for example) can be made.

13b. [2 marks]

Markscheme

(M1)

Note: Award (M1) for substitution of –1 into f (x)

= 0 (A1)(G2)

[2 marks]

Examiners report


time.




Candidates should:


31






In part (b) the answer could have been checked using the table on the GDC.

13c. [1 mark]

Markscheme

(0, –3) (A1)

OR

x = 0, y = –3 (A1)

Note: Award (A0) if brackets are omitted.

[1 mark]

Examiners report


time.




Candidates should:







In part (c) coordinates were required.

13d. [3 marks]

32

Markscheme

(A1)(A1)(A1)

Note: Award (A1) for each correct term. Award (A1)(A1)(A0) at most if there are extra terms.

[3 marks]

Examiners report


time.




Candidates should:







The responses to part (d) were generally correct.

13e. [1 mark]

Markscheme

(M1)

(AG)

Note: Award (M1) for substitution of x = –1 into correct derivative only. The final answer must be seen.

[1 mark]

Examiners report33


time.




Candidates should:







The “show that” nature of part (e) meant that the final answer had to be stated.

13f. [2 marks]

Markscheme

f '(–1) gives the gradient of the tangent to the curve at the point with x = –1. (A1)(A1)

Note: Award (A1) for “gradient (of curve)”, (A1) for “at the point with x = –1”. Accept “the

instantaneous rate of change of y” or “the (first) derivative”.

[2 marks]

Examiners report


time.




Candidates should:



34





The interpretive nature of part (f) was not understood by the majority.

13g. [2 marks]

Markscheme

(M1)

Note: Award (M1) for substituted in equation.

(A1)(G2)

Note: Accept y = –5.33x – 5.33.

OR

(M1)(A1)(G2)

Note: Award (M1) for substituted in equation, (A1) for correct equation. Follow through from

their answer to part (b). Accept y = –5.33 (x +1). Accept equivalent equations.

[2 marks]

Examiners report


time.

35




Candidates should:







13h. [2 marks]

Markscheme

(A1)(ft) for a tangent to their curve drawn.

(A1)(ft) for their tangent drawn at the point x = –1. (A1)(ft)(A1)(ft)

Note: Follow through from their graph. The tangent must be a straight line otherwise award at most

(A0)(A1).

[2 marks]

Examiners report


time.




Candidates should:





36



13i. [2 marks]

Markscheme

(i) (G1)

(ii) (G1)

Note: If a and b are reversed award (A0)(A1).

[2 marks]

Examiners report


time.




Candidates should:







Parts (i) and (j) had many candidates floundering; there were few good responses to these parts.

13j. [1 mark]

Markscheme

37

f (x) is increasing (A1)

[1 mark]

Examiners report


time.




Candidates should:







Parts (i) and (j) had many candidates floundering; there were few good responses to these parts.

14a. [4 marks]

Markscheme

(i) (M1)

OR

(M1)

Note: Award (M1) for setting the gradient function to zero.

(A1) (C2)

(ii) (M1)

38

(A1)(ft) (C2)

Note: Follow through from their .

[4 marks]

Examiners report

This question was not answered well at all except by the more able. Indeed, of the lower quartile of

candidates, the maximum mark achieved was only 1. Of those that did make a successful attempt at the

question, very few used the fact that preferring instead to differentiate and equate to

zero. But such candidates were in the minority as substituting into the given quadratic and

equating to zero produced the popular, but erroneous, answer of . Recovery was possible for the

next two marks if this incorrect value had been seen to be substituted into the correct quadratic, along

with to arrive at an answer of . This would have given (M1)(A1)(ft). However, candidates

who had an answer of in part (a)(i), invariably showed no working in part (ii) and

consequently earned no marks here. Irrespective of incorrect working in part (a), the quadratic

function clearly passes through (0, 4) and has a minimum at . Using this information, a

minority of candidates picked up at least one of the two marks in part (b).

14b. [2 marks]

Markscheme

39

(A1)(ft)(A1)(ft) (C2)

Notes: Award (A1)(ft) for a curve with correct concavity consistent with their passing through (0, 4).

(A1)(ft) for minimum in approximately the correct place. Follow through from their part (a).

[2 marks]

Examiners report

This question was not answered well at all except by the more able. Indeed, of the lower quartile of

candidates, the maximum mark achieved was only 1. Of those that did make a successful attempt at the

question, very few used the fact that preferring instead to differentiate and equate to

zero. But such candidates were in the minority as substituting into the given quadratic and

equating to zero produced the popular, but erroneous, answer of . Recovery was possible for the

next two marks if this incorrect value had been seen to be substituted into the correct quadratic, along

with to arrive at an answer of . This would have given (M1)(A1)(ft). However, candidates

who had an answer of in part (a)(i), invariably showed no working in part (ii) and

40

consequently earned no marks here. Irrespective of incorrect working in part (a), the quadratic

function clearly passes through (0, 4) and has a minimum at . Using this information, a

minority of candidates picked up at least one of the two marks in part (b).

15a. [2 marks]

Markscheme

(M1)

(A1)(G2)

[2 marks]

Examiners report

As usual and by intention, this question caused the most difficulty in terms of its content; however, for

those with a sound grasp of the topic, there were many very successful attempts. Much of the question

could have been answered successfully by using the GDC, however, it was also clear that a number of

candidates did not connect the question they were attempting with the curve that they had either

sketched or were viewing on their GDC. Where there was no alternative to using the calculus, many

candidates struggled.

The majority of sketches were drawn sloppily and with little attention to detail. Teachers must impress

on their students that a mathematical sketch is designed to illustrate the main points of a curve – the

smooth nature by which it changes, any symmetries (reflectional or rotational), positions of turning

points, intercepts with axes and the behaviour of a curve as it approaches an asymptote. There must

also be some indication of the dimensions used for the “window”.

Differentiation of terms with negative indices remains a testing process for the majority; it will

continue to be tested.

It was also evident that some centres do not teach the differential calculus.

15b. [4 marks]

Markscheme

41

(A1) for labels and some indication of scale in an appropriate window

(A1) for correct shape of the two unconnected and smooth branches

(A1) for maximum and minimum in approximately correct positions

(A1) for asymptotic behaviour at -axis (A4)

Notes: Please be rigorous.

The axes need not be drawn with a ruler.

The branches must be smooth: a single continuous line that does not deviate from its proper direction.

The position of the maximum and minimum points must be symmetrical about the origin.

The -axis must be an asymptote for both branches. Neither branch should touch the axis nor must the

curve approach the

asymptote then deviate away later.

[4 marks]

Examiners report







42









15c. [3 marks]

Markscheme

(A1)(A1)(A1)

Notes: Award (A1) for , (A1) for , (A1) for . Award a maximum of (A1)(A1)(A0) if extra

terms seen.

[3 marks]

Examiners report














43


15d. [2 marks]

Markscheme

(M1)

Note: Award (M1) for substitution of into their derivative.

(A1)(ft)(G1)

[2 marks]

Examiners report















15e. [2 marks]

Markscheme

or , (G1)(G1)

44

Notes: Award (G0)(G0) for , . Award at most (G0)(G1) if parentheses are omitted.

[2 marks]

Examiners report















15f. [3 marks]

Markscheme

(A1)(A1)(ft)(A1)(ft)

Notes: Award (A1)(ft) or seen, (A1)(ft) for or , (A1) for weak (non-

strict) inequalities used in both of the above.

Accept use of in place of . Accept alternative interval notation.

Follow through from their (a) and (e).

If domain is given award (A0)(A0)(A0).

Award (A0)(A1)(ft)(A1)(ft) for , .

Award (A0)(A1)(ft)(A1)(ft) for , .

45

[3 marks]

Examiners report















15g. [2 marks]

Markscheme

(M1)(A1)(ft)(G2)

Notes: Award (M1) for seen or substitution of into their derivative. Follow through from

their derivative if working is seen.

[2 marks]

Examiners report





46











15h. [2 marks]

Markscheme

(M1)(A1)(ft)(G2)

Notes: Award (M1) for equating their derivative to their or for seeing parallel lines on their graph

in the approximately correct position.

[2 marks]

Examiners report












47




16a. [2 marks]

Markscheme

(A1)(A1)

Note: Award (A1) for , (A1) for .

[2 marks]

Examiners report

Part a) was either answered well or poorly.

16b. [3 marks]

Markscheme

(A1)(A1)(A1)

Notes: Award (A1) for , (A1) for , (A1) for . Award (A1)(A1)(A0) at most if any other term

present.

[3 marks]

Examiners report

Most candidates found the first term of the derivative in part b) correctly, but the rest of the terms were

incorrect.

16c. [2 marks]

Markscheme

(M1)

(A1)(ft)(G2)

Note: Follow through from their derivative function.

48

[2 marks]

Examiners report

The gradient in c) was for the most part correctly calculated, although some candidates substituted

incorrectly in instead of in .

16d. [2 marks]

Markscheme

Decreasing, the derivative (gradient or slope) is negative (at ) (A1)(R1)(ft)

Notes: Do not award (A1)(R0). Follow through from their answer to part (c).

[2 marks]

Examiners report

Part d) had mixed responses.

16e. [4 marks]

Markscheme

49

(A4)

Notes: Award (A1) for labels and some indication of scales and an appropriate window.

Award (A1) for correct shape of the two unconnected, and smooth branches.

Award (A1) for the maximum and minimum points in the approximately correct positions.

Award (A1) for correct asymptotic behaviour at .

Notes: Please be rigorous.

The axes need not be drawn with a ruler.

The branches must be smooth and single continuous lines that do not deviate from their proper

direction.

The max and min points must be symmetrical about point .

The -axis must be an asymptote for both branches.

[4 marks]

Examiners report

50

Lack of labels of the axes, appropriate scale, window, incorrect maximum and minimum and incorrect

asymptotic behaviour were the main problems with the sketches in e).

16f. [4 marks]

Markscheme

(i) or , (G1)(G1)

(ii) or , (G1)(G1)

[4 marks]

Examiners report

Part f) was also either answered correctly or entirely incorrectly. Some candidates used the trace

function on the GDC instead of the min and max functions, and thus acquired coordinates with

unacceptable accuracy. Some were unclear that a point of local maximum may be positioned on the

coordinate system “below” the point of local minimum, and exchanged the pairs of coordinates of those

points in f(i) and f(ii).

16g. [3 marks]

Markscheme

or (A1)(A1)(ft)(A1)

Notes: (A1)(ft) for or . (A1)(ft) for or . (A1) for weak (non-strict)

inequalities used in both of the above. Follow through from their (e) and (f).

[3 marks]

Examiners report

Very few candidates were able to identify the range of the function in (g) irrespective of whether or not

they had the sketches drawn correctly.

17a. [2 marks]

Markscheme51

x = 0, x = 4 (A1)(A1) (C2)

Notes: Accept 0 and 4.

[2 marks]

Examiners report

A number of candidates left out this question which indicated that this topic was either entirely

unfamiliar, that this topic of the syllabus had perhaps not been taught, or was barely familiar. A few

candidates wrote down coordinate pairs when asked for a solution to the equation. A number of

candidates wrote down the formula for the equation of the axis of symmetry without being able to

substitute values for a and b. When given the minimum value of the graph a small number of candidates

could identify the range of the function correctly. Overall this question proved to be difficult with its

demands for reading and interpreting the graph, and dealing with additional information about the

quadratic function given in the different parts.

17b. [2 marks]

Markscheme

x = 2 (A1)(A1) (C2)

Note: Award (A1) for x = constant, (A1) for 2.

[2 marks]

Examiners report







52



17c. [1 mark]

Markscheme

x = –2 (A1) (C1)

Note: Accept –2.

[1 mark]

Examiners report









17d. [1 mark]

Markscheme

(A1) (C1)

Notes: Accept alternative notations.

Award (A0) for use of strict inequality.

[1 mark]

Examiners report

53









18a. [2 marks]

Markscheme

4a + 2b = 20

a + b = 8 (A1)

a – b = –4 (A1) (C2)

Note: Award (A1)(A1) for any two of the given or equivalent equations.

[2 marks]

Examiners report

Most candidates attempted this question but very few of them completed it entirely. A number of

students wrote incorrect equations in part (a), which shows that the mapping diagram was poorly

understood and read. Part (c) proved to be difficult for many who didn’t know how to find the x-

coordinate of the vertex of the graph of the function. Some students gave the two coordinates instead of

the x-coordinate only.

18b. [1 mark]

Markscheme

a = 2 (A1)(ft)

[1 mark]

Examiners report

54






18c. [1 mark]

Markscheme

b = 6 (A1)(ft) (C2)

Note: Follow through from their (a).

[1 mark]

Examiners report

Most candidates attempted this question but very few of them completed it entirely. A numberof





18d. [2 marks]

Markscheme

(M1)

Note: Award (M1) for correct substitution in correct formula.

(A1)(ft) (C2)

[2 marks]

Examiners report55






19a. [1 mark]

Markscheme

q = 4 (A1) (C1)

[1 mark]

Examiners report

This question was not well answered with few candidates gaining full marks. Many candidates could

find the value of q but not r. Although many found the minimum value of y, they could not find the

maximum value of the function or express the range correctly.

19b. [2 marks]

Markscheme

(M1)

r = 10 (A1) (C2)

[2 marks]

Examiners report




19c. [1 mark]

Markscheme

–8.5 (A1)(ft) (C1)

[1 mark]

56

Examiners report




19d. [2 marks]

Markscheme

(A1)(ft)(A1)(ft) (C2)

Notes: Award (A1)(ft) for their answer to part (c) with correct inequality signs, (A1)(ft) for 104.

Follow through from their values of q and r.

Accept 104 ±2 if read from graph.

[2 marks]

Examiners report




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57

ib questionbank test€¦ · web view[1 mark] marks. cheme (a1) (c1) notes: accept alternative...

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